### Solving K_{sp} Problems: Part One - x^{2}

Part Two - 4x^{3}

Part Three - 27x^{4}

Part Four - 108x^{5}

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There are five substances I will use as examples. I will do AgCl in this file and do two others in each of part two and part three.

The problem will always be the same in each example. It is:

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

Silver chloride, AgCl, has a K_{sp} = 1.77 x 10¯^{10}. Calculate its solubility in moles per liter.
The dissociation equation is as before:

AgCl (s) <===> Ag^{+} (aq) + Cl¯ (aq)

The K_{sp} expression is as before:

K_{sp} = [Ag^{+}] [Cl¯]

This is the equation we must solve. First we put in the K_{sp} value:

1.77 x 10¯^{10} = [Ag^{+}] [Cl¯]

Now, we have to reason out the values of the two guys on the right, First of all, I have no idea what the values are so I'll use unknowns. Like this:

[Ag^{+}] = x

[Cl¯] = y

However, I now have a BIG problem (No, not B.I.G., just BIG). The problem is that I have two unknowns, but only one equation. So it seems I'm stuck, untillllll . . .

I examine the chemical equation and see that there is a one-to-one ratio between Ag^{+} and Cl¯. I know this from the coefficients (both one) of the balanced equation.

That means that the concentrations of the two ions are EQUAL. I can use the same unknown to represent both. Like this:

[Ag^{+}] = x = [Cl¯]

Substituting, we get:

1.77 x 10¯^{10} = (x) (x)

Now, we take the square root of both sides. I hope I'm not too insulting when I emphasize both sides. I have had lots of people take the square root of the x^{2} side, but not the other. After the square root, we get:

x = 1.33 x 10¯^{5} M

This is the answer because there is a one-to-one relationship between the Ag^{+} dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯^{5} moles per liter.

One last thing. The K_{sp} value does not have any units on it, but when you get to the value for x, be sure to put M (for molarity) on it. The reasons behind this are complex and beyond the scope of the ChemTeam's chosen area.

Do you want to see more problems like this?

Go to Solving K_{sp} Problems - Part Two

Go to Solving K_{sp} Problems - Part Three

Go to Solving K_{sp} Problems - Part Four

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